I realized there was a way to solve this in a single line (excluding the import) by using the triangular number formula. There's no real advantage here, except that we don't need to count the pairs separately:

Instead, we count the occurrences n for each number and calculate the number of pairs from 0…n-1

from collections import Counter


class Solution:
  def numGoodPairs(self, nums):
    return sum((v * (v - 1) // 2) for _, v in Counter(nums).items())